# Differentiable almost every where

I need help in this problem ,royden 4th edition . Let $g$ be integrable over $[a,b]$ .Define the antiderivative of $g$ to be the function $f$ defined on $[a,b]$ by $f(x)=\int_{a}^{x} g~~ for~all~ x\in [a,b]$. Show that $f$ is differentiable almost everywhere on $(a,b)$.

I know from Lebesgues Theorem if the function $f$ is monotone on the open intervals $(a,b)$ , then it is differentiable almost everywhere on $(a,b)$ But integrable not always monotone .

and I can not show that if $f$ is not differentiable at $x$ then $x \in E$ and $m(E)=0$ These are the only information I have about differentiable almost everywhere

• For real-valued $g$, write $f$ as the difference of two monotonic functions. $g = \lvert g\rvert - (\lvert g\rvert - g)$ – Daniel Fischer Oct 28 '15 at 19:25
• @DanielFischer Why should $g$ equal such a difference? – zhw. Oct 28 '15 at 19:41
• Whether $g$ is Lebesgue integrable or Riemann integrable we can write $g=g_1-g_2$ where $g_1$ and $g_2$ are nonnegative and integrable in the same sense. If eva had known that functions of bounded variation are a.e. differentiable then no hint would be needed. – B. S. Thomson Oct 28 '15 at 19:55
Since $$g = |{g}| - (|{g}|-g)$$ and $$g$$ is integrable, the linearity of Lebesgue integral implies \begin{align*} f(x) = \int_a^x g = \int_a^x |{g}| - (|{g}|-g) = \int_a^x |{g}| - \int_a^x(|{g}|-g) = f_1(x)-f_2(x), \end{align*} where $$f_1(x) = \int_a^x |{g}|$$ and $$f_2(x) = \int_a^x(|{g}|-g)$$. Since $$|{g}|\geq 0$$ and $$|{g}|-g\geq0$$, the functions $$f_1$$ and $$f_2$$ are increasing on $$(a,b)$$. Thus, $$f_1$$ and $$f_2$$ are differentiable almost every on $$(a,b)$$. Since addition preserves differentiability, $$f$$ is differentiable a.e. on $$(a,b)$$.